Lineintegral $int_{gamma}|z|^2dz$ over ellipse












1














Let $a,binmathbb{R}_{>0}$ and $gamma: [0,2pi]rightarrowmathbb{C},tmapsto acos(t)+ibsin(t)$



calculate the line integral



$int_{gamma}|z|^2dz$



My calculation turns out to be really ugly. Is there maybe a "nice" way to calculate this integral?






integration complex-analysis complex-integration line-integrals







share|cite|improve this question



























    1














    Let $a,binmathbb{R}_{>0}$ and $gamma: [0,2pi]rightarrowmathbb{C},tmapsto acos(t)+ibsin(t)$



    calculate the line integral



    $int_{gamma}|z|^2dz$



    My calculation turns out to be really ugly. Is there maybe a "nice" way to calculate this integral?






    integration complex-analysis complex-integration line-integrals







    share|cite|improve this question

























      1












      1








      1







      Let $a,binmathbb{R}_{>0}$ and $gamma: [0,2pi]rightarrowmathbb{C},tmapsto acos(t)+ibsin(t)$



      calculate the line integral



      $int_{gamma}|z|^2dz$



      My calculation turns out to be really ugly. Is there maybe a "nice" way to calculate this integral?






      integration complex-analysis complex-integration line-integrals







      share|cite|improve this question













      Let $a,binmathbb{R}_{>0}$ and $gamma: [0,2pi]rightarrowmathbb{C},tmapsto acos(t)+ibsin(t)$



      calculate the line integral



      $int_{gamma}|z|^2dz$



      My calculation turns out to be really ugly. Is there maybe a "nice" way to calculate this integral?






      integration complex-analysis complex-integration line-integrals




      integration complex-analysis complex-integration line-integrals






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 20 '18 at 15:49









      Christian Singer

      334213




      334213






















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          The integrals are not ugly at all. You obtain
          $$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
          whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.






          share|cite|improve this answer





















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            1 Answer
            1






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            oldest

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            active

            oldest

            votes









            3














            The integrals are not ugly at all. You obtain
            $$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
            whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.






            share|cite|improve this answer


























              3














              The integrals are not ugly at all. You obtain
              $$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
              whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.






              share|cite|improve this answer
























                3












                3








                3






                The integrals are not ugly at all. You obtain
                $$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
                whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.






                share|cite|improve this answer












                The integrals are not ugly at all. You obtain
                $$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
                whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 '18 at 16:22









                Christian Blatter

                172k7112326




                172k7112326






























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